2. Revision and Consolidation 08/09/2002By Mr. NGAN HON SHING1 Revision and Consolidation

3. 08/09/2002By Mr. NGAN HON SHING2 Newton ’ s Law  Newton ’ s 1 st Law  Newton ’ s 2 nd Law The net force on an object  rate of change of momentum of an object. F  dp/dt => dp/dt = d(mv)/dt = m(dv/dt)+v(dm/dt) = ma where k=1  Newton ’ s 3 rd Law Action and reaction are equal but opposite. They will not cancel each other. Show 3 forces on diagram. 1 balanced forces and 1 action-reaction pair.

4. Revision and Consolidation 08/09/2002By Mr. NGAN HON SHING3 Newton ’ s 3 rd Law (A-R pair) W (earth -> block) R (floor -> block) A (block -> floor) Balanced Forces: W and R – acting on the same object A-R pair: A and R – acting on 2 different objects. They will not cancel each other.

5. Revision and Consolidation 08/09/2002By Mr. NGAN HON SHING4 Conservation of linear momentum It ’ s true for all collision if there is no external force acting on the colliding bodies. m 1 u 1 +m 2 u 2 = m 1 v 1 +m 2 v 2 Types of collisions: Elastic collision (perfectly/completely) No energy lost (Partially) Inelastic collision Energy lost Completely inelastic collision Energy lost is max.

6. Revision and Consolidation 08/09/2002By Mr. NGAN HON SHING5 Conservation of linear momentum We cannot use the above equation to identify what type of collision it is. Conservation of mechanical energy: If the above eqt is true, it is an elastic collision.

7. Revision and Consolidation 08/09/2002By Mr. NGAN HON SHING6 Energy (work done, PE, KE & P) Work done = W.D. = +ve => energy gained W.D. = -ve => energy lost KE = PE = gravitational potential energy = mgh P = rate of change of energy = = Fv

8. Revision and Consolidation 08/09/2002By Mr. NGAN HON SHING7 Derive KE in terms of m and v WD by F = F. s = (ma) s = u=0 => Wd by F = ½ mv 2 s F u=0 v